Complexity Explorer Santa Fe Institute

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Vector and Matrix Algebra

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2.6 Eigenvalues and Eigenvectors » Quiz #15 Solution

Question 1:  Recall that to find the eigenvalues \lambda of A, we solve the following equation:

det(A-I\lambda)=0

det (A-I\lambda)=det \left(\begin{bmatrix} 2 & 1 \\ 0 & -1 \end{bmatrix}- \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} \right)= det \begin{bmatrix} 2 - \lambda & 1 \\ 0 & -1 -\lambda \end{bmatrix} = (2-\lambda)(-1-\lambda)-1=\lambda^2-\lambda-2

=(\lambda-2)(\lambda+1)=0

Thus, the eigenvalues of A are  2, -1.

 

Question 2:

det(B-I\lambda)=0

det (B-I\lambda)=det \left(\begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix}- \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} \right)= det \begin{bmatrix} 1 - \lambda & -2 \\ 1 & 4 -\lambda \end{bmatrix} = (1-\lambda)(4-\lambda)+2=\lambda^2-5\lambda +6

=(\lambda-2)(\lambda-3)=0

Thus, the eigenvalues of A are  2,3.